The equation of a circle $C$ is $x^2+y^2-18x+18y+161 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-18x) + (y^2+18y) = -161$ $(x^2-18x+81) + (y^2+18y+81) = -161 + 81 + 81$ $(x-9)^{2} + (y+9)^{2} = 1 = 1^2$ Thus, $(h, k) = (9, -9)$ and $r = 1$.